Thread: Differentiable function

1. Differentiable function

As h->0, show that(explain why)
sin(x+h) = sinx +hcosx + o(h)

2. some thing like this always depends on what you already know and are allowed to use.

If you know that the derivative of sin x is cos x, then you can find the tangent line to sin x at $\displaystyle x= -x_0$:

$\displaystyle sin(x)= cos(x_0)(x- x_0)+ sin(x_0)$

Taking $\displaystyle x= x_0+ h$, $\displaystyle sin(x_0+ h)= cos(x_0)(h)+ sin(x_0)$ and, replacing $\displaystyle x_0$ with x,
sin(x+ h)= cos(x)h+ sin(x).

Another way would be to use the Taylor's series for sin(x) and cos(x):
$\displaystyle sin(x)= x- \frac{x^2}{2!}+ \frac{x^4}{4!}+ \cdot\cdot\cdot$
$\displaystyle cos(x)= 1- \frac{x^3}{3!}+ \frac{x^5}{5!}+ \dot\cdot\cdot$

3. Could you explain this part? I'm not sure how you got the formula from just knowing the derivative of sin.. and how you got the tangent line at sinx at $\displaystyle x= -x_0$
Originally Posted by HallsofIvy
If you know that the derivative of sin x is cos x, then you can find the tangent line to sin x at $\displaystyle x= -x_0$:

$\displaystyle sin(x)= cos(x_0)(x- x_0)+ sin(x_0)$

4. Originally Posted by Monster32432421
Could you explain this part? I'm not sure how you got the formula from just knowing the derivative of sin.. and how you got the tangent line at sinx at $\displaystyle x= -x_0$
Normally, one of the first things people learn in Calculus is that the derivative is the slope of the tangent line and that, if y= f(x), the tangent line at $\displaystyle (x_0, f(x_0))$ is $\displaystyle y= f'(x_0)(x- x_0)+ f(x_0)$. Check your notes.

5. Originally Posted by HallsofIvy
Normally, one of the first things people learn in Calculus is that the derivative is the slope of the tangent line and that, if y= f(x), the tangent line at $\displaystyle (x_0, f(x_0))$ is $\displaystyle y= f'(x_0)(x- x_0)+ f(x_0)$. Check your notes.
Another question...How come you are missing the o(h) in the final solution?

thanks so much for your help so far..

6. Sorry person who is apparently not that awesome at maths.

"The above equation is wrong. The function on the LHS is the sine
function, whereas on the RHS you have a straight line. How can they
equal?

The tangent line is NOT equal to the curve. They only meet at ONE point:
the touching point."

Could someone else do my question?....

7. Do you know that if $\displaystyle \lim_{x\to a} f(x) = L$ then $\displaystyle f(x) = L + o(1)$ where the o(1) is as $\displaystyle x\to a$? Write down the difference quotient and use this to turn it into the tangent line estimate $\displaystyle f(x+h) = f(x) + f'(x) h +o(h)$, then set f(x) = sin(x).

Also, be nice.

edit: Or were you trying to prove that the derivative of sine is the cosine?